\(\int \frac {x^2}{(b x^{2/3}+a x)^{3/2}} \, dx\) [175]

Optimal result

Integrand size = 19, antiderivative size = 160 \[ \int \frac {x^2}{\left (b x^{2/3}+a x\right )^{3/2}} \, dx=-\frac {6 x^2}{a \sqrt {b x^{2/3}+a x}}-\frac {256 b^3 \sqrt {b x^{2/3}+a x}}{21 a^5}+\frac {512 b^4 \sqrt {b x^{2/3}+a x}}{21 a^6 \sqrt [3]{x}}+\frac {64 b^2 \sqrt [3]{x} \sqrt {b x^{2/3}+a x}}{7 a^4}-\frac {160 b x^{2/3} \sqrt {b x^{2/3}+a x}}{21 a^3}+\frac {20 x \sqrt {b x^{2/3}+a x}}{3 a^2} \] Output:

-6*x^2/a/(b*x^(2/3)+a*x)^(1/2)-256/21*b^3*(b*x^(2/3)+a*x)^(1/2)/a^5+512/21 
*b^4*(b*x^(2/3)+a*x)^(1/2)/a^6/x^(1/3)+64/7*b^2*x^(1/3)*(b*x^(2/3)+a*x)^(1 
/2)/a^4-160/21*b*x^(2/3)*(b*x^(2/3)+a*x)^(1/2)/a^3+20/3*x*(b*x^(2/3)+a*x)^ 
(1/2)/a^2
 

Mathematica [A] (verified)

Time = 4.87 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.53 \[ \int \frac {x^2}{\left (b x^{2/3}+a x\right )^{3/2}} \, dx=\frac {512 b^5 \sqrt [3]{x}+256 a b^4 x^{2/3}-64 a^2 b^3 x+32 a^3 b^2 x^{4/3}-20 a^4 b x^{5/3}+14 a^5 x^2}{21 a^6 \sqrt {b x^{2/3}+a x}} \] Input:

Integrate[x^2/(b*x^(2/3) + a*x)^(3/2),x]
 

Output:

(512*b^5*x^(1/3) + 256*a*b^4*x^(2/3) - 64*a^2*b^3*x + 32*a^3*b^2*x^(4/3) - 
 20*a^4*b*x^(5/3) + 14*a^5*x^2)/(21*a^6*Sqrt[b*x^(2/3) + a*x])
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {1921, 1922, 1922, 1922, 1908, 1920}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^2}{\left (a x+b x^{2/3}\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1921

\(\displaystyle \frac {10 \int \frac {x}{\sqrt {x^{2/3} b+a x}}dx}{a}-\frac {6 x^2}{a \sqrt {a x+b x^{2/3}}}\)

\(\Big \downarrow \) 1922

\(\displaystyle \frac {10 \left (\frac {2 x \sqrt {a x+b x^{2/3}}}{3 a}-\frac {8 b \int \frac {x^{2/3}}{\sqrt {x^{2/3} b+a x}}dx}{9 a}\right )}{a}-\frac {6 x^2}{a \sqrt {a x+b x^{2/3}}}\)

\(\Big \downarrow \) 1922

\(\displaystyle \frac {10 \left (\frac {2 x \sqrt {a x+b x^{2/3}}}{3 a}-\frac {8 b \left (\frac {6 x^{2/3} \sqrt {a x+b x^{2/3}}}{7 a}-\frac {6 b \int \frac {\sqrt [3]{x}}{\sqrt {x^{2/3} b+a x}}dx}{7 a}\right )}{9 a}\right )}{a}-\frac {6 x^2}{a \sqrt {a x+b x^{2/3}}}\)

\(\Big \downarrow \) 1922

\(\displaystyle \frac {10 \left (\frac {2 x \sqrt {a x+b x^{2/3}}}{3 a}-\frac {8 b \left (\frac {6 x^{2/3} \sqrt {a x+b x^{2/3}}}{7 a}-\frac {6 b \left (\frac {6 \sqrt [3]{x} \sqrt {a x+b x^{2/3}}}{5 a}-\frac {4 b \int \frac {1}{\sqrt {x^{2/3} b+a x}}dx}{5 a}\right )}{7 a}\right )}{9 a}\right )}{a}-\frac {6 x^2}{a \sqrt {a x+b x^{2/3}}}\)

\(\Big \downarrow \) 1908

\(\displaystyle \frac {10 \left (\frac {2 x \sqrt {a x+b x^{2/3}}}{3 a}-\frac {8 b \left (\frac {6 x^{2/3} \sqrt {a x+b x^{2/3}}}{7 a}-\frac {6 b \left (\frac {6 \sqrt [3]{x} \sqrt {a x+b x^{2/3}}}{5 a}-\frac {4 b \left (\frac {2 \sqrt {a x+b x^{2/3}}}{a}-\frac {2 b \int \frac {1}{\sqrt [3]{x} \sqrt {x^{2/3} b+a x}}dx}{3 a}\right )}{5 a}\right )}{7 a}\right )}{9 a}\right )}{a}-\frac {6 x^2}{a \sqrt {a x+b x^{2/3}}}\)

\(\Big \downarrow \) 1920

\(\displaystyle \frac {10 \left (\frac {2 x \sqrt {a x+b x^{2/3}}}{3 a}-\frac {8 b \left (\frac {6 x^{2/3} \sqrt {a x+b x^{2/3}}}{7 a}-\frac {6 b \left (\frac {6 \sqrt [3]{x} \sqrt {a x+b x^{2/3}}}{5 a}-\frac {4 b \left (\frac {2 \sqrt {a x+b x^{2/3}}}{a}-\frac {4 b \sqrt {a x+b x^{2/3}}}{a^2 \sqrt [3]{x}}\right )}{5 a}\right )}{7 a}\right )}{9 a}\right )}{a}-\frac {6 x^2}{a \sqrt {a x+b x^{2/3}}}\)

Input:

Int[x^2/(b*x^(2/3) + a*x)^(3/2),x]
 

Output:

(-6*x^2)/(a*Sqrt[b*x^(2/3) + a*x]) + (10*((2*x*Sqrt[b*x^(2/3) + a*x])/(3*a 
) - (8*b*((6*x^(2/3)*Sqrt[b*x^(2/3) + a*x])/(7*a) - (6*b*((6*x^(1/3)*Sqrt[ 
b*x^(2/3) + a*x])/(5*a) - (4*b*((2*Sqrt[b*x^(2/3) + a*x])/a - (4*b*Sqrt[b* 
x^(2/3) + a*x])/(a^2*x^(1/3))))/(5*a)))/(7*a)))/(9*a)))/a
 

Defintions of rubi rules used

rule 1908

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + 
b*x^n)^(p + 1)/(a*(j*p + 1)*x^(j - 1)), x] - Simp[b*((n*p + n - j + 1)/(a*( 
j*p + 1)))   Int[x^(n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, j, n, 
p}, x] &&  !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(n*p + n - j + 1)/(n 
- j)], 0] && NeQ[j*p + 1, 0]
 

rule 1920

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[(-c^(j - 1))*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j 
)*(p + 1))), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[ 
n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])
 

rule 1921

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[(-c^(j - 1))*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j 
)*(p + 1))), x] + Simp[c^j*((m + n*p + n - j + 1)/(a*(n - j)*(p + 1)))   In 
t[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, j, m, n} 
, x] &&  !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/( 
n - j)], 0] && LtQ[p, -1] && (IntegerQ[j] || GtQ[c, 0])
 

rule 1922

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^(j - 1)*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p 
+ 1))), x] - Simp[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1)))   I 
nt[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, 
p}, x] &&  !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(m + n*p + n - j + 1) 
/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c, 0])
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.48

method	result	size
derivativedivides	\(\frac {2 x \left (x^{\frac {1}{3}} a +b \right ) \left (7 a^{5} x^{\frac {5}{3}}-10 x^{\frac {4}{3}} a^{4} b +16 a^{3} b^{2} x -32 a^{2} b^{3} x^{\frac {2}{3}}+128 x^{\frac {1}{3}} a \,b^{4}+256 b^{5}\right )}{21 \left (b \,x^{\frac {2}{3}}+a x \right )^{\frac {3}{2}} a^{6}}\)	\(77\)
default	\(\frac {2 x \left (x^{\frac {1}{3}} a +b \right ) \left (7 a^{5} x^{\frac {5}{3}}-10 x^{\frac {4}{3}} a^{4} b +16 a^{3} b^{2} x -32 a^{2} b^{3} x^{\frac {2}{3}}+128 x^{\frac {1}{3}} a \,b^{4}+256 b^{5}\right )}{21 \left (b \,x^{\frac {2}{3}}+a x \right )^{\frac {3}{2}} a^{6}}\)	\(77\)

Input:

int(x^2/(b*x^(2/3)+a*x)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

2/21*x*(x^(1/3)*a+b)*(7*a^5*x^(5/3)-10*x^(4/3)*a^4*b+16*a^3*b^2*x-32*a^2*b 
^3*x^(2/3)+128*x^(1/3)*a*b^4+256*b^5)/(b*x^(2/3)+a*x)^(3/2)/a^6
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1598 vs. \(2 (120) = 240\).

Time = 114.81 (sec) , antiderivative size = 1598, normalized size of antiderivative = 9.99 \[ \int \frac {x^2}{\left (b x^{2/3}+a x\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(x^2/(b*x^(2/3)+a*x)^(3/2),x, algorithm="fricas")
 

Output:

-1/21*((3145728*a^3*b^13 + 2621440*a^3*b^12 - 983040*a^3*b^11 - 10192*a^12 
 + 196608*(17*a^6 - 3*a^3)*b^10 + 4096*(464*a^6 + 53*a^3)*b^9 - 6144*(246* 
a^6 + a^3)*b^8 + 768*(1120*a^9 - 2560*a^6 - 3*a^3)*b^7 - 256*(548*a^9 - 15 
69*a^6)*b^6 - 768*(1477*a^9 + 31*a^6)*b^5 - 48*(2304*a^12 + 21176*a^9 + 33 
*a^6)*b^4 - 4032*(96*a^12 - 23*a^9)*b^3 - 12*(27648*a^12 + 527*a^9)*b^2 + 
3*(39296*a^12 + 51*a^9)*b)*x^2 + (3145728*b^16 + 2621440*b^15 + 196608*(17 
*a^3 - 3)*b^13 - 983040*b^14 + 4096*(464*a^3 + 53)*b^12 - 10192*a^9*b^3 - 
6144*(246*a^3 + 1)*b^11 + 768*(1120*a^6 - 2560*a^3 - 3)*b^10 - 256*(548*a^ 
6 - 1569*a^3)*b^9 - 768*(1477*a^6 + 31*a^3)*b^8 - 48*(2304*a^9 + 21176*a^6 
 + 33*a^3)*b^7 - 4032*(96*a^9 - 23*a^6)*b^6 - 12*(27648*a^9 + 527*a^6)*b^5 
 + 3*(39296*a^9 + 51*a^6)*b^4)*x - 2*(7*(4096*a^7*b^9 + 6144*a^7*b^8 + 768 
*a^7*b^7 - 4096*a^13 - 144*a^10*b^2 + 216*a^10*b - 27*a^10 + 256*(16*a^10 
- 7*a^7)*b^6 + 48*(128*a^10 - 3*a^7)*b^5 + 24*(32*a^10 + 9*a^7)*b^4 - (588 
8*a^10 + 27*a^7)*b^3)*x^3 - 58*(4096*a^4*b^12 + 6144*a^4*b^11 + 768*a^4*b^ 
10 - 144*a^7*b^5 + 216*a^7*b^4 + 256*(16*a^7 - 7*a^4)*b^9 + 48*(128*a^7 - 
3*a^4)*b^8 + 24*(32*a^7 + 9*a^4)*b^7 - (5888*a^7 + 27*a^4)*b^6 - (4096*a^1 
0 + 27*a^7)*b^3)*x^2 - 128*(4096*a*b^15 + 6144*a*b^14 + 768*a*b^13 + 256*( 
16*a^4 - 7*a)*b^12 - 144*a^4*b^8 + 48*(128*a^4 - 3*a)*b^11 + 216*a^4*b^7 + 
 24*(32*a^4 + 9*a)*b^10 - (5888*a^4 + 27*a)*b^9 - (4096*a^7 + 27*a^4)*b^6) 
*x + (1048576*b^16 + 1572864*b^15 + 65536*(16*a^3 - 7)*b^13 + 196608*b^...
 

Sympy [F]

\[ \int \frac {x^2}{\left (b x^{2/3}+a x\right )^{3/2}} \, dx=\int \frac {x^{2}}{\left (a x + b x^{\frac {2}{3}}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(x**2/(b*x**(2/3)+a*x)**(3/2),x)
 

Output:

Integral(x**2/(a*x + b*x**(2/3))**(3/2), x)
 

Maxima [F]

\[ \int \frac {x^2}{\left (b x^{2/3}+a x\right )^{3/2}} \, dx=\int { \frac {x^{2}}{{\left (a x + b x^{\frac {2}{3}}\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^2/(b*x^(2/3)+a*x)^(3/2),x, algorithm="maxima")
 

Output:

integrate(x^2/(a*x + b*x^(2/3))^(3/2), x)
 

Giac [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.70 \[ \int \frac {x^2}{\left (b x^{2/3}+a x\right )^{3/2}} \, dx=-\frac {512 \, b^{\frac {9}{2}}}{21 \, a^{6}} + \frac {6 \, b^{5}}{\sqrt {a x^{\frac {1}{3}} + b} a^{6}} + \frac {2 \, {\left (7 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {9}{2}} a^{48} - 45 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {7}{2}} a^{48} b + 126 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {5}{2}} a^{48} b^{2} - 210 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {3}{2}} a^{48} b^{3} + 315 \, \sqrt {a x^{\frac {1}{3}} + b} a^{48} b^{4}\right )}}{21 \, a^{54}} \] Input:

integrate(x^2/(b*x^(2/3)+a*x)^(3/2),x, algorithm="giac")
 

Output:

-512/21*b^(9/2)/a^6 + 6*b^5/(sqrt(a*x^(1/3) + b)*a^6) + 2/21*(7*(a*x^(1/3) 
 + b)^(9/2)*a^48 - 45*(a*x^(1/3) + b)^(7/2)*a^48*b + 126*(a*x^(1/3) + b)^( 
5/2)*a^48*b^2 - 210*(a*x^(1/3) + b)^(3/2)*a^48*b^3 + 315*sqrt(a*x^(1/3) + 
b)*a^48*b^4)/a^54
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\left (b x^{2/3}+a x\right )^{3/2}} \, dx=\int \frac {x^2}{{\left (a\,x+b\,x^{2/3}\right )}^{3/2}} \,d x \] Input:

int(x^2/(a*x + b*x^(2/3))^(3/2),x)
 

Output:

int(x^2/(a*x + b*x^(2/3))^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.42 \[ \int \frac {x^2}{\left (b x^{2/3}+a x\right )^{3/2}} \, dx=\frac {\frac {2 x^{\frac {5}{3}} a^{5}}{3}-\frac {64 x^{\frac {2}{3}} a^{2} b^{3}}{21}-\frac {20 x^{\frac {4}{3}} a^{4} b}{21}+\frac {256 x^{\frac {1}{3}} a \,b^{4}}{21}+\frac {32 a^{3} b^{2} x}{21}+\frac {512 b^{5}}{21}}{\sqrt {x^{\frac {1}{3}} a +b}\, a^{6}} \] Input:

int(x^2/(b*x^(2/3)+a*x)^(3/2),x)
 

Output:

(2*(7*x**(2/3)*a**5*x - 32*x**(2/3)*a**2*b**3 - 10*x**(1/3)*a**4*b*x + 128 
*x**(1/3)*a*b**4 + 16*a**3*b**2*x + 256*b**5))/(21*sqrt(x**(1/3)*a + b)*a* 
*6)